27-30 Find the temperature distribution in a laterally = 1.158 cm²/sec), insulated thin copper bar (c² = K/po K/po 50 cm long and of constant cross section with endpoints at x = 0 and 50 kept at 0°C and initial temperature
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- H.W 1. Consider a two degree of freedom bar elements as shown in figure. Using finite element method to formulate the equilibrium equation of it. If the cross sectional area is 12 mm and E=200 GN/m². 20KN +30KN 15KN 300 * 600- + 350 All Dimension in mm 2. Consider a two degree of freedom bar element as shown in figure. Using finite element method to formulate the equilibrium equation of it, and then estimate the stress distributions. If Esteel-200 GN/m, Ecopper 110GN/m and EAL= 120 GN/m?. d=3 Steel AL Соpper 20KN - 30KN →15KN 300 * 200 - 400 * 350 All Dimension in mmA very thick structure is subjected to certain traction boundary conditions on its surface. The cross-section and the applied load do not vary with the z-coordinate. The following stress function is proposed for this problem: -y p(x,y) = Sin (x) (A x²e + B e") (i) use the biharmonic equation to find restrictions, if any, on values of A and B (ii) calculate all stress components (iii) calculate all strain components in terms of A, B, and C as well as the Young modulus and Poisson's ratio E and y, respectively. (iv) check that the equilibrium equations are satisfied (v) determine the traction boundary conditions at x =± a and y=+b= 73,0 GPa As = 1500 mm ² ) rod and a brass (Eg = 100 G+Pa t are stress-free and and AB = 2000mm²) rod. At 25°C, The rods of 1.00mm exosts between them. дар expansion If the coefficrear of thermal as of = 14 [106 m/m'c] that of brass a f = 11 [10° ⁰ m/m²c], A Temperature increased to 350°C.. a A Stech | E₁ kuhat is intermidate Temperature (umt°C) that system must be taken to in order to clase the of 1.00mm? A it Steel 1.00mm → K B Brass 1.50000m. for steal is taken as and 1.0 ME the gap 15
- Determine the strain energy per unit volume (units of psi) and the strain energy per unit weight (units of in ) that can be stored in each or the materials listed in the accompanying table, assuming that the material is stressed to the proportional limit. DATA FOR PROBLEM 2.7-5 Material Weight Density (lb/in3) Modulus of Elasticity (ksi) Proportional Limit (psi) Mild sleel 0.284 30,000 36,000 Tool steel 0.284 30,000 75,000 Aluminum 0.0984 10,500 60,000 Rubber (soft) 0.0405 0.300 300Problem 2. A block of cast iron was wrapped in the thin aluminum foil as shown on the figure. At room temperature there is no stress on the foil. The whole system is then immersed in liquid nitrogen at -196°C. Find the stress on the aluminum foil. Is the foil going to deform permanently? Is it going to break? Stress [MPa] 350 300 250 200 150 100 50 0 0 alu alloy 5083-H34 0.005 0.01 0.015 0.02 Strain Aluminum foil 5 µm thick, this is a continuous strip with the ends welded together Cast iron 20 mm thick a aluminum = 2.4 105 K-¹, Stress MPa 400 300 200 100 a iron 1.2 105 K-1 ↑ 11 0 0 0.05 0.1 0.15 0.2 0.25 Strain Fig. 2. Stress-strain diagrams for aluminum. Left: Low strain region. Right: Full strain range. Hint: you can get the Young's modulus, proportions limit (elastic limit), and break point from the graphs.1.4-7 The data shown in the table below were obtained from a tensile test of high-strength steel. The test specimen had a diameter of 13 mm and a gage length of 50 mm (see figure for Prob. 1.4-3). At fracture, the elongation between the gage marks was 3.0 mm and the minimum diameter was 10.7 mm. Plot the conventional stress-strain curve for the steefor the steel and determine the proportional limit, modulus of elastics of elastic- ity (i.e., the slope of the initial part of the stress-strain,tress-strain curve), yield stress at 0.1% offset, ultimate stress, percent, elongation in 50 mm, and percent reduction in area. 'ess, percent area. TENSILE-TEST DATA FOR PROB. 1.4-7 Elongation (mm) 0.005 0.015 0.048 Load (kN) 5 10 30 50 0.084 60 0.099 64.5 0.109 67.0 0.119 68.0 0.137 69.0 0.160 70.0 0.229 72.0 0.259 76.0 0.330 84.0 0.584 92.0 0.853 100.0 1.288 112.0 2.814 113.0 Fracture
- By the equipment as shown in the figure, temperature can be measured. The bar AB and CD are made from aluminum and titanium. The coefficients of thermal expansion are 23.6x106 1/K and 8.2x10-6 1/k. Obtain the relationship between the temperature and the vertical displacement at the point E. 6 mm 72 mm 36 mm DFigure below shows the bar with three equal elements. Use the finite element method and calculate: 2.1 the global stiffness matrix 2.2 the displacement on node 2, 3, 4 2.3 the Strain in each element 2.4 the stresses in each element using Hook's law and compare with theoretical stresses (o=F/A) (1) (2) (3) -50 N 10 mm 10 mm 10 mm A1=50 mm? A2=20 mm? A:=10 mm? E=200 GPaA metal wire has the following properties. Length Im, Initial temperature= 80sin(rz100), The ends kept zero, Density = 9.36g/cm3, Specific heat 0,125cal/gC, Thermal conductivity = 1.34cal/(cmseeC) Find the temperature at 20cm from the first edge of the bar at t 85 u(20, 85)
- Where T1>T2>T3, Illustrate how attractive or repulsive force (which causes non-ideal behavior) affects the isotherm at one of the temperatures. Also indicate the corresponding compression factor Z values * Create an axes that show x-axis labeled as 1/V (K) and y-axis labeled as P1 Consider a two degree of freedom bar elements as shown in figure. Using finite element method to formulate the equilibrium equation of it. If the cross sectional area is 12 mm and E=200 GN/m2. 12 20KN 30KN 1SKN 300 600*350 bom All Dimensioon in numIf the displacements applied to a 3D isotropic material element is defined as functions of (x,y,z), Determine the following: 1- Stress at point (2,3,4). Assume E = 12.08 MPa and v = 0.25 2- Strain at point (1,2,2) 3- If you consider a plane stress condition for the above problems show how will the 2D assumption change the stress at the same point